Question

The total number of instantaneous centres for a mechanism of \( n \) links are:

• A. \( ^nC_2 \)
• B. \( ^nP_2 \)
• C. \( ^nC_2 + ^nP_2 \)
• D. \( ^nC_2 - ^nP_2 \)

Hint:

Remember the formula for combinations \( ^nC_r \) as it directly applies to finding the number of pairs of links in a mechanism. Each pair of links corresponds to one instantaneous center.

Answer 1

The Correct Option is A

Step 1: Understand the concept of instantaneous centers.
An instantaneous center (also known as centro or virtual center) is a point in a mechanism about which a rigid link is instantaneously rotating relative to another rigid link. According to Aronhold Kennedy's Theorem, for a planar mechanism having \( n \) links, there are \( ^nC_2 = \frac{n(n-1)}{2} \) instantaneous centers of rotation. Step 2: Recall Aronhold Kennedy's Theorem.
Aronhold Kennedy's Theorem states that for a system of three rigid bodies having planar motion, the three instantaneous centers must lie on a straight line. This theorem is fundamental in locating the instantaneous centers in a mechanism. Step 3: Determine the number of instantaneous centers.
Consider a mechanism with \( n \) links. Each pair of links has a relative motion, and thus, there exists an instantaneous center of rotation for each pair. The number of ways to choose 2 links from \( n \) links is given by the combination formula \( ^nC_2 \). The formula for combinations is: $$^nC_r = \frac{n!}{r!(n-r)!}$$In our case, we want to choose 2 links from \( n \) links, so \( r = 2 \):$$^nC_2 = \frac{n!}{2!(n-2)!} = \frac{n(n-1)(n-2)!}{2 \times 1 \times (n-2)!} = \frac{n(n-1)}{2}$$ Therefore, the total number of instantaneous centers for a mechanism of \( n \) links is \( ^nC_2 \).